Improved bounds on the size of the smallest representation of relation algebra $32_{65}$

Abstract

In this paper, we shed new light on the spectrum of the relation algebra we call $A_{n}$, which is obtained by splitting the non-flexible diversity atom of $6_{7}$ into n symmetric atoms. Precisely, show that the minimum value in $\text {Spec}(A_{n})$ is at most $2n^{6 + o(1)}$, which is the first polynomial bound and improves upon the previous bound due to Dodd and Hirsch (J Relat Methods Comput Sci 2:18–26, 2013). We also improve the lower bound to $2n^{2} + 4n + 1$, which is roughly double the trivial bound of $n^{2} + 2n + 3$. In the process, we obtain stronger results regarding $\text {Spec}(A_{2}) =\text {Spec}(32_{65})$. Namely, we show that 1024 is in the spectrum, and no number smaller than 26 is in the spectrum. Our improved lower bounds were obtained by employing a SAT solver, which suggests that such tools may be more generally useful in obtaining representation results.

1 Introduction

Relation algebra $32_{65}$ has atoms $1'$, a, b, and c, all symmetric, with all diversity cycles not involving a forbidden. The atom a is flexible, and $32_{65}$ has the mandatory cycles required to make a flexible and no others. The numbering system for finite integral relation algebras is due to Maddux [7].

Relation algebra $32_{65}$ was shown in [1] to be representable over a finite set, namely a set of 416,714,805,914 points. This was reduced in [5] to 63,432,274,896 points, which was later reduced to 8192 by the first and sixth authors (unpublished), and finally to 3432 in [2]. Here, we give the smallest known representation, over 1024 points.

There are few published lower bounds in the literature. Most can be found in [4], where the spectrum of every relation algebra with three or fewer atoms is determined. Going up to four atoms increases the difficulty considerably.

No lower bound on the size of representations of $32_{65}$ has been published. We give a non-trivial such bound for an infinite class of algebras in Section 3.

The importance of the relation algebras studied here, namely $32_{65}$ and those derived from splitting a non-flexible diversity atom of $32_{65}$, stems from the flexible atom conjecture:

Conjecture 1.1

(Flexible Atom Conjecture). Every finite integral relation algebra with a flexible atom has a representation over a finite set.

In [6], the finite symmetric integral relation algebras in which every diversity atom is flexible were shown to be finitely representable. In particular, the algebra with n flexible atoms is representable over a set of size $(2+o(1))n^2$. In the present paper, we look at the other extreme, where only one atom is flexible, and the only mandatory cycles are those that involve the flexible atom, i.e., only those that are required to make the atom flexible. We show that in this case as well, the number of points required grows only polynomially in the number of atoms. (See Theorem 2.3.)

Now we give the requisite definitions.

Definition 1.2

A relation algebra is an algebra $\langle A, +, -, ;, \breve{}, 1'\rangle$ such that

• $\langle A, +, -\rangle$ is a Boolean algebra

• $\langle A, ;, \breve{}, 1'\rangle$ is an involuted monoid

• $x \mathbin {;} (y+z) = x\mathbin {;} y + x\mathbin {;} z$

• $(x+y)\;\breve{} = \breve{x} + \breve{y}$

• For all x, y, and z, we have

  $$\begin{aligned} x\mathbin {;}y \cdot \breve{z} =0 \iff y\mathbin {;}z \cdot \breve{x} =0. \end{aligned}$$

Definition 1.3

For a relation algebra A, any $a \in A$ is an atom if $a \ne 0$ and $b < a$ implies $b=0$; it is called a diversity atom if, in addition, $a\cdot 1' = 0$.

Definition 1.4

For diversity atoms a, b, c, the triple (a, b, c), usually written abc, is called a diversity cycle. A cycle abc is called forbidden if $a\mathbin {;}b \cdot c = 0$ and mandatory if $a\mathbin {;}b \ge c$. (Note that these are the only possibilities, otherwise c is not an atom.)

Definition 1.5

We say that a symmetric diversity atom f is flexible if for all diversity atoms a, b, we have that abf is mandatory.

Definition 1.6

A relation algebra A is called representable if there is a set U and an equivalence relation $E\subseteq U\times U$ such that A embeds in

$$\begin{aligned} \langle \mathrm {Powerset}(E), \cup ,\ ^c, \circ , ^{-1}, Id_E\rangle . \end{aligned}$$

In this paper, we will be concerned only with simple RAs, so E can be equal to $U\times U$ for some set U. A representation where $E = U\times U$ is called square.

Definition 1.7

Let A be a finite relation algebra. Then

$$\begin{aligned} {{\,\mathrm{Spec}\,}}(A)=\{\alpha \le \omega : A\text { has a square representation over a set of size } \alpha \}. \end{aligned}$$

Let $A_n$ denote the integral symmetric relation algebra with atoms $1'$, r, $b_1$, ..., $b_n$, where a diversity cycle is mandatory if and only if it involves the atom r. (So $A_2$ is $32_{65}$.) Let

$$\begin{aligned} f(n) = \min ( {{\,\mathrm{Spec}\,}}(A_n)). \end{aligned}$$

It was shown in [1] that f(n) is finite for all n.

Because representing finite integral relation algebras amounts to edge-coloring complete graphs with the diversity atoms, we will use the language of graph theory. So that we can use colors to make pretty pictures, we will refer to a as red, to b as light blue, and to c as dark blue.

2 An upper bound on f(n)

In this section, we give a representation of $32_{65}$ over 1024 points, and then generalize to give representations of $A_n$ for all n.

Consider $G=(\mathbb {Z}/2\mathbb {Z})^{10}$, and consider the elements as bitstrings. Define

$$\begin{aligned} R&=\{x \in G : x \text { has between one and six 1s} \}, \text { and } \\ B&=\{x \in G : x \text { has at least seven 1s} \}. \end{aligned}$$

This defines a group representation of $6_7$, which is a subalgebra of $32_{65}$. There exists a way of splitting B into $B_1$ and $B_2$ so that:

• $R+B_i=G\smallsetminus \{0\}$, $i=1,2$;

• $B_i+B_i = R \cup \{0\}$, $i=1,2$;

• $B_1+B_2 = R$.

This yields a group representation of $32_{65}$ over $2^{10}=1024$ points, improving the previous smallest-known representation over ${14 \atopwithdelims ()7}=3432$ points [2]. We note that while the representation given here is smaller, the representation over 3432 points in [2] has a nice, compact description.

The split was found in the following way. The first author checked several million random splits. None of them worked, but some got “close”. He took one of the close ones and tinkered with it for about three hours until it worked. The curious can view the process in the Jupyter notebook 32_65 splitting.ipynb at https://github.com/algorithmachine/RA-32-of-65.

The following Python 3 code can be used to verify that the given split yields a representation. (Bitstrings are encoded as integers between 0 and 1023. Note that the Python operator $\wedge$ denotes the bitwise exclusive-or operation, which is the group operation in our setting.)

We generalize this argument as follows:

Theorem 2.1

For all $n\ge 2$, $A_n$ is representable over $(\mathbb {Z}/2\mathbb {Z})^{3k+1}$ for sufficiently large $k\in \omega$. In particular, for $n\ge 14$, it suffices to take $k=n$.

Remark 2.2

Theorem 2.1 tells us that f(n) is at most exponential in n. In contrast, the most you could say from [1] was that f(n) was bounded above by (roughly) ${15n^2 \atopwithdelims ()n}$. See Figure 1.

Fig. 1

Upper bound on f(n) vs. n

Proof

We actually prove something more general: we split both the flexible atom and the non-flexible atom into n parts, since we get this stronger result with essentially no more work. We have already shown that for $k = 3$, $A_{2}$ can be realized over $(\mathbb {Z}/2\mathbb {Z})^{3k+1}$. We now argue that for $n \ge 3$, $A_{n}$ can be realized over $(\mathbb {Z}/2\mathbb {Z})^{3k+1}$ for sufficiently large $k \in \omega$.

Our approach is to use the probabilistic method to show that, given a large enough representation of the relation algebra $6_{7}$ over $(\mathbb {Z}/2\mathbb {Z})^{3k+1}$, the atom b can be partitioned into n parts, as $A_{n}$ is obtained from $6_{7}$ by splitting, as in [3].

Consider $G = (\mathbb {Z}/2\mathbb {Z})^{3k+1}$. For $x\in G$, let x(i) denote the $i^\text {th}$ coordinate of x. Let |x| denote $|\{ i:x(i)=1\} |$. Denote $\text {supp}(x) = \{ i : x(i) = 1\}$ to be the support of x. The key idea is the following partition of $G\smallsetminus \{ 0\}$ into two sets R and B.

Let:

$$\begin{aligned}&R=\{ x\in G:1\le |x|\le 2k\}\quad \text {and } \\&B=\{ x\in G: 2k+1\le |x|\le 3k+1\}. \end{aligned}$$

Then $R+R=G$, $R+B=G\smallsetminus \{ 0\}$, and $B+B=R\cup \{0\}$. As we will see below, B is a sum-free set with high additive energy.

We now split both the “red” and “blue” atoms of $6_7$ into n atoms and find a representation over a finite set. Namely, we split R and B into n parts $R_1,\ldots ,R_n$ and $B_1,\ldots ,B_n$ uniformly at random. We need to count the “witnesses” to the “needs” of each element. We will show that each need is witnessed at least $2^k$ times. Consider the following cases.

• Case 1: We count witnesses for $R\subseteq B+B$. Let $z \in R$, and denote $\ell := |z|$. We consider two sub-cases: whether $1 \le \ell \le k$, and whether $k+1 \le \ell \le 2k$.

  • Case 1.1: Suppose first that $1\le \ell \le k$. We construct x, y randomly so that $z=x+y$. For each $i\in \text {supp}(z)$, we choose uniformly at random whether:

    $$\begin{aligned} x(i)=1&\quad \text {and}\quad y(i)=0\\&\text {OR}\\ x(i)=0&\quad \text {and}\quad y(i)=1. \end{aligned}$$

    As $|z| = \ell$, this yields $2^\ell$ possible selections. For the $k-\ell$ left-most indices $j\notin \text {supp}(z)$, we choose uniformly at random whether:

    $$\begin{aligned} x(j)&=1=y(j)\\&\text {OR}\\ x(j)&=0 =y(j). \end{aligned}$$

    As there are $k-\ell$ positions, there are $2^{k-\ell }$ possible selections. For the remaining $2k+1$ positions i, we let $x(i)=1=y(i)$. Thus, we have that $x, y \in B$. By the rule of product, we obtain $2^{\ell } \cdot 2^{k-\ell } = 2^{k}$ possible selections. It follows that there are at least $2^{k}$ witnesses for z.

  • Case 1.2: Suppose now that $k+1\le \ell \le 2k$. For the k least indices $i\in \text {supp}(z)$, we choose uniformly at random whether:

    $$\begin{aligned} x(i)=1&\quad \text {and}\quad y(i)=0\\&\text {OR}\\ x(i)=0&\quad \text {and}\quad y(i)=1 \end{aligned}$$

    For the remaining $\ell -k$ indices $i\in \text {supp}(z)$, let $x(i)=1$ and $y(i)=0$, or $x(i)=0$ and $y(i)=1$ in such a way that ensures that both x and y have at least $\ell -k$ 1’s in coordinates in $\text {supp}(z)$. Then for all indices $j\notin \text {supp}(z)$, let $x(j)=1=y(j)$. There were k flips, so there are at least $2^k$ witnesses.

  It follows that if $z \in R$, there are at least $2^{k}$ ways to witness z as the sum $x+y$, where $x, y \in B$.

• Case 2: Now let us consider witnesses to $B\subseteq B+R$. Let $z \in B$, and denote $\ell := |z|$. By the definition of B, we have that $2k+1 \le \ell \le 3k.$ We randomly construct $x\in B$, $y\in R$ so that $z=x+y$. For the $2k+1$ indices $i\in \text {supp}(z)$ of least index, set $x(i)=1$ and $y(i)=0$. For the remaining $\ell -(2k+1)$ indices $i\in \text {supp}(z)$, we choose uniformly at random whether:

  $$\begin{aligned} x(i)=1&\quad \text {and}\quad y(i)=0\\&\text {OR}\\ x(i)=0&\quad \text {and}\quad y(i)=1. \end{aligned}$$

  For each index $j\notin \text {supp}(z)$, we choose uniformly at random whether:

  $$\begin{aligned} x(j)&=1=y(j)\\&\text {OR}\\ x(j)&=0 =y(j). \end{aligned}$$

  Again, there were k flips, so we have at least $2^k$ witnesses.

• Case 3: Next, let us consider witnesses to $B\subseteq R+R$. Let $z\in B$. We construct $x,y\in R$ so that $z = x+y$. For every $j\not \in \text {supp}(z)$, set $x(j)=0=y(j)$. This is to ensure that $x, y \in R$. For the smallest k indices $i\in \text {supp}(z)$, we choose uniformly at random whether:

  $$\begin{aligned} x(i)=1&\text { and }y(i)=0\\&\text {OR}\\ x(i)=0&\text { and } y(i)=1. \end{aligned}$$

  For the remaining indices $i\in \text {supp}(z)$, we choose uniformly at random whether:

  $$\begin{aligned} x(i)=1&\text { and }y(i)=0\\&\text {OR}\\ x(i)=0&\text { and } y(i)=1 \end{aligned}$$

  in such a way that ensures that neither x nor y receives more than 2k 1’s. Clearly, there are at least $2^k$ witnesses.

• Case 4: Now we consider witnesses for $R\subseteq B+R$. Let $z\in R$, and denote $\ell := |z|$. We build $x\in B$, $y\in R$ so that $z=x+y$. We consider the following sub-cases, namely whether $1 \le \ell \le k$, and whether $k+1 \le \ell \le 2k$.

  • Case 4.1: First, consider the case where $1\le \ell \le k$. For $i \in \text {supp}(z)$, set $x(i)=1$ and $y(i)=0$. For $j\notin \text {supp}(z)$, choose $2k+1-\ell$ of the $3k+1-\ell$ indices, and set $x(j)=1=y(j)$. Set all others to $x(j)=0=y(j)$. Since ${3k+1-\ell \atopwithdelims ()2k+1-\ell } \ge {2k\atopwithdelims ()k}>2^k$, we have at least $2^k$ witnesses.

  • Case 4.2: Now consider the case where $k+1\le n\le 2k$. For the smallest k indices $i\in \text {supp}(z)$, set $x(i)=1$ and $y(i)=0$. For the remaining $\ell -k$ indices $i\in \text {supp}(z)$, we choose uniformly at random whether:

    $$\begin{aligned} x(i)=1&\text { and }y(j)=0\\&\text {OR}\\ x(i)=0&\text { and } y(j)=1. \end{aligned}$$

    Now we choose $k+1$ of the remaining $3k+1-\ell$ indices $j\notin \text {supp}(z)$. There are ${3k+1-\ell \atopwithdelims ()k+1}$ choices, which ranges between ${k+1\atopwithdelims ()k+1}$ and ${2k+1\atopwithdelims ()k+1}$. Therefore there are at least $2^{\ell -k}\cdot {3k+1-\ell \atopwithdelims ()k+1}$ witnesses. It is not hard to check that for $0\le N\le k$, ${k+1+N\atopwithdelims ()k+1}>2^N$, and therefore $2^{\ell -k}\cdot {3k+1-\ell \atopwithdelims ()k+1}>2^{\ell -k}\cdot 2^{2k-\ell }=2^k$.

• Case 5: Finally, we consider witnesses for $R\subseteq R+R$. Let $z\in R$, and denote $\ell := |z|$. We construct $x,y\in R$ so that $z=x+y$. We consider the following cases: whether $1 \le \ell \le k$, and whether $k+1 \le \ell \le 2k$.

  • Case 5.1: First, consider the case where $1\le \ell \le k$. For each $i\in \text {supp}(z)$, set $x(i)=1$ and $y(i)=0$. Then for the smallest 2k indices outside of $\text {supp}(z)$, choose k of then. For each such selected j, set $x(j)=1=y(j)$, and $x(j)=0=y(j)$ otherwise. This gives at least ${2k\atopwithdelims ()k}>2^k$ witnesses.

  • Case 5.2: We next consider the case where $k+1\le \ell \le 2k$. For each $i \in \text {supp}(z)$, we choose uniformly at random whether:

    $$\begin{aligned} x(i)=1&\text { and }y(i)=0\\&\text {OR}\\ x(i)=0&\text { and } y(i)=1. \end{aligned}$$

    This gives $2^\ell >2^k$ witnesses.

Now we are ready to compute the probability that our random partition $R_1,\ldots ,R_n$ and $B_1,\ldots ,B_n$ fails to be a representation. Let $z\in (\mathbb {Z}/2\mathbb {Z})^{3k+1} \smallsetminus \{{\textbf {0}}\}$. If $z\in R_i$, then z has $3n^2$ “needs”:

• $\forall i,j \ z\in R_i+R_j$

• $\forall i,j \ z\in R_i+B_j$

• $\forall i,j \ z\in B_i+B_j$.

If $z\in B_j$, then z has $2n^2$ “needs”:

• $\forall i,j \ z\in R_i+R_j$

• $\forall i,j \ z\in R_i+B_j$.

So $3n^2$ is a bound on the number of “needs”. Fix z, and let $x, y \in G$ such that $x + y = z$. The probability that the edge xy witnesses a fixed need is $1/n^{2}$. So the probability that the edge xy does not witness a fixed need is $(1 - 1/n^{2})$. For a particular need, there are at least $2^{k}$ witnesses. As we color the edges uniformly at random, the probability that a fixed need of z is unsatisfied is at most:

$$\begin{aligned} \left( 1 - \frac{1}{n^{2}}\right) ^{2^{k}}. \end{aligned}$$

As z has at most $3n^{2}$ needs, we have that

$$\begin{aligned} Pr[z\text { has an unsatisfied need}] \le 3n^2\left( 1-\frac{1}{n^2}\right) ^{2^k}. \end{aligned}$$

Thus

$$\begin{aligned} Pr[\exists z\text { with an unsatisfied need}]&\le \sum _z 3n^2\left( 1-\frac{1}{n^2}\right) ^{2^k}\end{aligned}$$

(2.1)

$$\begin{aligned}&=2^{3k+1}\cdot 3n^2\left( 1-\frac{1}{n^2}\right) ^{2^k} \end{aligned}$$

(2.2)

$$\begin{aligned}&\le 2^{3(k+1)}\cdot n^2\left( 1-\frac{1}{n^2}\right) ^{2^k}. \end{aligned}$$

(2.3)

We want (2.3) to be less than 1, which is equivalent to its logarithm being less than zero:

$$\begin{aligned}&\log \left( 2^{3(k+1)}\cdot n^2\cdot \left( 1-\frac{1}{n^2}\right) ^{2^k}\right)<0 \\&\quad \Longleftrightarrow 3(k+1)\log 2+2\log n+2^k\log \left( \frac{n^2-1}{n^2}\right)<0\\&\quad \Longleftrightarrow 3(k+1)\log 2+2\log n<2^k\log \left( \frac{n^2}{n^2-1}\right) \end{aligned}$$

Now assuming $3 \le n \le k$, we have that

$$\begin{aligned} 3(k+1)\log 2+2\log n&<3(k+1)+k\\&\le 5k. \end{aligned}$$

Thus

$$\begin{aligned} 2^k\cdot \log \left( \frac{n^2}{n^2-1}\right)&=2^k[\log (n^2)-\log (n^2-1)]\\&>2^k\cdot \frac{1}{n^2}, \end{aligned}$$

where the last inequality is due to the fact that $\log (t+1)-\log (t) < 1/t$, which follow the concavity of $\log$. So we need $5k<\frac{2^k}{n^2}$. Setting $k=n$, we have $5n^3<2^n$, which holds for all $n\ge 14$. Hence taking $k=n$ gives a non-zero probability that a random partition yields a representation. $\square$

The construction in the proof of Theorem 2.1 provides the bound $f(n) \le 2^{3n+1}$ for $n \ge 14$. By fine-tuning our choice of k, we obtain polynomial bounds on f(n). Let $k = (2+o(1))\log (n)$. We have by (2.2) that

$$\begin{aligned} Pr[\exists z\text { with an unsatisfied }\text {need}]&\le 2^{3k+1}\cdot 3n^2\left( 1-\frac{1}{n^2}\right) ^{2^k} \\&= 2^{3(2+o(1))\log (n) + 1} \cdot 3n^{2} \cdot \left( 1-\frac{1}{n^2}\right) ^{2^{(2+o(1))\log (n)}} \\&= 2n^{6 + o(1)} \cdot 3n^{2} \cdot \left( 1-\frac{1}{n^2}\right) ^{2^{(2+o(1))\log (n)}} \\&= 6n^{8+o(1)} \cdot \left( 1-\frac{1}{n^2}\right) ^{2^{(2+o(1))\log (n)}}. \end{aligned}$$

Now we have that

$$\begin{aligned} \lim _{n \rightarrow \infty } 6n^{8+o(1)} \cdot \left( 1-\frac{1}{n^2}\right) ^{2^{(2+o(1))\log (n)}} = 0. \end{aligned}$$

(2.4)

So choosing $k = (2+o(1))\log (n)$ yields the following.

Theorem 2.3

For n sufficiently large, we have that $f(n) \le 2n^{6 + o(1)}$.

We note that the threshold $n_{0}$ for which Theorem 2.3 applies is quite large. For instance, choosing $k = 3\log (n)$ yields that $f(n) \le 2n^{9}$, which holds for all $n \ge 91$. So in choosing $k = (2+o(1))\log (n)$, the bound of $f(n) \le 2n^{6+o(1)}$ holds for all $n \ge n_{0} \ge 91$, where $n_{0}$ depends on o(1). This contrasts with the bound in Theorem 2.1, which holds for all $n \ge 14$. Furthermore, calibrating our choice of $k = c \log (n)$ failed to yield improvements on $f(3) \le 2^{16}$ and $f(4) \le 2^{19}.$

It is natural to ask whether modifying our choice of k in this construction will yield additional improvements in the upper bound for f(n). If we take $k = 2\log (n)$ rather than $k = (2+o(1))\log (n)$, we have that

$$\begin{aligned} \lim _{n \rightarrow \infty } 6n^{8} \cdot \left( 1-\frac{1}{n^2}\right) ^{2^{2\log (n)}} = \infty . \end{aligned}$$

The key reason behind this is that

$$\begin{aligned} \lim _{n \rightarrow \infty } \left( 1-\frac{1}{n^2}\right) ^{2^{2\log (n)}} = \lim _{n \rightarrow \infty } \left( 1-\frac{1}{n^2}\right) ^{n^{2}} = \frac{1}{e}. \end{aligned}$$

This suggests that further analyzing the Boolean cube is unlikely to yield additional improvements on the upper bound for f(n).

We also note that (2.4) yields that as $n \rightarrow \infty$, the probability that there exists $z \in (\mathbb {Z}/2\mathbb {Z})^{3k+1}$ with an unsatisfied need goes to 0. So with high probability, a random partition yields a representation of $A_n$. We record this observation with the following corollary.

Corollary 2.4

Suppose that we split both the “red” and “blue” atoms of $6_7$ into n atoms, as in the proof of Theorem 2.1. Namely, we split R and B into n parts $R_1,\ldots ,R_n$ and $B_1,\ldots ,B_n$ uniformly at random. With high probability, we have that such a random split is a representation of a relation algebra containing a subalgebra isomorphic to $A_n$.

3 A lower bound

In this section, we consider representations of $32_{65}$ as edge-colorings of $K_n$ with all mandatory triangles present and no all-blue triangles. Note that blue triangles are forbidden even if they contain edges of differing shades of blue. In other words, every triangle must contain a red edge.

We now make our representation precise. Let $\rho :32_{65} \rightarrow \text {Powerset}(U\times U)$, where $U = \{x_0, \dots , x_{n-1}\}$, be a representation. Then label the vertices of $K_n$ with $\{x_0, \dots , x_{n-1}\}$, and let the color of edge $x_ix_j$ be the atom z such that $(x_i, x_j) \in \rho (z)$.

Lemma 3.1

${{\,\mathrm{Spec}\,}}(32_{65}) \subseteq \{11, \ldots \} \cup \{\omega \}$.

Proof

There must be some red edge $x_0x_1$. Any red edge has nine “needs”. There must be nine points that witness these needs, which together with $x_0$ and $x_1$ make a total of 11 points. (See Figure 2.) $\square$

Fig. 2

The needs of a red edge

We can easily obtain a slight improvement using the classical Ramsey number R(4, 3).

Lemma 3.2

$11 \not \in {{\,\mathrm{Spec}\,}}(32_{65})$.

Proof

We know that at least 11 points are required. Since $R(4,3)=9$, and there are no all-blue triangles, there must be a red $K_4$. Let $x_0x_1$ be an edge in this red $K_4$. Then $x_0x_1$ must have its red-red need met twice, hence there must be ten points besides $x_0$ and $x_1$. $\square$

Lemma 3.3

In any representation of $32_{65}$, for every red edge there is a red $K_4$ that is vertex-disjoint from it. In particular, off of every red edge $x_0 x_1$ one can find the configuration depicted in Figure 3.

Fig. 3

Subgraph which must appear off of any red edge

Proof

Let $x_0x_1$ be red, with witnesses to all needs as in Figure 2. Then $\{x_2,x_3,x_4,x_5\}$ induce a red $K_4$, since any blue edge among them would create an all-blue triangle with $x_0$ (and also with $x_1$). Furthermore, any edge running from any of $x_2,x_3,x_4,x_5$ to any of $x_6,x_7,x_8,x_9$ must be red, since any such blue edge would create an all-blue triangle with either $x_0$ (for $x_7$ and $x_9$) or $x_1$ (for $x_6$ and $x_8$). Thus we have the configuration depicted in Figure 3. $\square$

Lemma 3.4

$12,13,14,15,16 \not \in {{\,\mathrm{Spec}\,}}(32_{65})$.

Proof

Consider the configuration depicted in Figure 3. The edge $x_2x_5$ is red. Then $x_0$ and $x_1$ both witness the light-blue-dark-blue need, while $x_3,x_4,x_6,$ $x_7,x_8$ and $x_9$ all witness the red-red need. There are seven needs yet unsatisfied. The remaining vertex $x_{10}$ could witness some need, but vertices $x_{11}$ through $x_{16}$ will have to be added. Thus there are at least 17 points. See Figure 4. $\square$

Fig. 4

witnesses to the needs of $x_2x_5$

Lemma 3.4 generalizes nicely as follows.

Theorem 3.5

For all n, $f(n) \ge 2n^2+4n+1$.

Note that the trivial bound is $n^2+2n+3$, roughly half the bound in Theorem 3.5.

Proof

Call the shades of blue $b_1$ through $b_n$. Fix a red edge $x_0x_1$. Let BB denote the set of vertices that witness a blue-blue need for $x_0x_1$, and let RB denote the set of vertices that witness either a red-blue need or a blue-red need for $x_0x_1$. BB induces a red clique, and all edges from BB to RB are red. Note that $|BB|=n^2$ and $|RB|=2n$. This gives the trivial lower bound of $n^2+2n+3$.

Let $u\in BB$ witness $b_1$-$b_1$ for $x_0x_1$ and let $v\in BB$ witness $b_2$-$b_2$ for $x_0x_1$. The edge uv is red, hence has $(n+1)^2$ needs. Both $x_0$ and $x_1$ witness the same $b_1$-$b_2$ need, and all points in BB and RB (besides u and v) witness the red-red need. Hence there must be at least $(n+1)^2 - 2$ points outside of $\{x_0,x_1\}\cup BB \cup RB$. Hence there are at least $2 + n^2 + 2n + (n+1)^2 - 2 = 2n^2 + 4n + 1$ points. $\square$

Remark 3.6

Note that if two points u, v satisfy a red-blue need for $x_{0}x_{1}$, then uv is necessarily red. Otherwise, $uvx_{1}$ would form a blue triangle. There are n points that satisfy the red-blue need for $x_{0}x_{1}$. As the points in BB form a red clique of size $n^{2}$, we obtain the following.

Corollary 3.7

In any representation of $A_n$, the clique number of the red subgraph of the underlying graph of the representation is at least $n^2 + n$.

4 SAT solver results

In this section, we improve the lower bound on f(2) using a SAT solver.

Lemma 4.1

$17 \not \in {{\,\mathrm{Spec}\,}}(32_{65})$.

Proof

We build an unsatisfiable boolean formula $\Phi$ whose satisfiability is a necessary condition for $32_{65}$ to be representable over 17 points. For all $0\le i< j < 17$ and $k=0,1,2$, define a boolean $\phi _{i,j,k}$. We interpret $\phi _{i,j,0}$ being TRUE to mean that $x_ix_j$ is red, $\phi _{i,j,1}$ being TRUE to mean that $x_ix_j$ is light blue, and $\phi _{i,j,2}$ being TRUE to mean that $x_ix_j$ is dark blue. Then define

$$\begin{aligned} \Phi _0&= \bigwedge _{i<j} [(\phi _{i,j,0} \vee \phi _{i,j,1} \vee \phi _{i,j,2}) \wedge (\lnot \phi _{i,j,0} \vee \lnot \phi _{i,j,1}) \\&\quad \wedge (\lnot \phi _{i,j,0} \vee \lnot \phi _{i,j,2}) \wedge (\lnot \phi _{i,j,1} \vee \lnot \phi _{i,j,2}) ] \end{aligned}$$

Then $\Phi _0$ asserts that for each $i<j$, exactly one of $\phi _{i,j,0}$, $\phi _{i,j,1}$, and $\phi _{i,j,2}$ is TRUE.

Consider the subgraph depicted in Figure 3. Let

• $R = \{(i,j): i<j,\ x_ix_j \text { is red}\}$

• $Bl = \{(i,j): i<j,\ x_ix_j \text { is light blue}\}$

• $Bd = \{(i,j): i<j,\ x_ix_j \text { is dark blue}\}$

Define

$$\begin{aligned} \Phi _1 = \left( \bigwedge _{(i,j)\in R} \phi _{i,j,0}\right) \wedge \left( \bigwedge _{(i,j)\in Bl} \phi _{i,j,1}\right) \wedge \left( \bigwedge _{(i,j)\in Bd} \phi _{i,j,2} \right) \end{aligned}$$

Then $\Phi _1$ asserts that any edges colored in Figure 3 are colored correctly.

Finally, define

$$\begin{aligned} \Phi _2 = \bigwedge _{(i,j)\in R}\left[ \bigwedge _{c_1,c_2\in \{0,1,2\}} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \end{aligned}$$

Then $\Phi _2$ asserts that every red edge in Figure 3 has its needs satisfied.

Let $\Phi = \Phi _0 \wedge \Phi _1 \wedge \Phi _2$.

$\Phi$ has been verified by SAT solver to be unsatisfiable when there are 17 points. $\square$

Remark 4.2

From the previous lemma, we see that at least 18 points are required to represent $32_{65}$. Since $R(6,3)=18$, there must be a red $K_6$. We also obtain a red $K_{6}$ from Corollary 3.7, using the fact that $A_{2} = 32_{65}$. Thus we can always find the subgraph depicted in Figure 5.

Fig. 5

Mandatory subgraph with red $K_6$

Unfortunately, $\Phi$ is satisfiable on 18 or more points. We must add more clauses to make $\Phi$ unsatisfiable. First, expand R to include the red $K_6$ as in Figure 5. Second, we add clauses to forbid all-blue triangles:

$$\begin{aligned} \Phi _3 = \bigwedge _{i<j<k} ( \phi _{i,j,0} \vee \phi _{i,k,0} \vee \phi _{j,k,0} ) \end{aligned}$$

Lemma 4.3

Let $\Phi = \Phi _0 \wedge \Phi _1 \wedge \Phi _2\wedge \Phi _3$. Then for $n=18$, 19, $\Phi$ is unsatisfiable. Hence $18,19 \not \in {{\,\mathrm{Spec}\,}}(32_{65})$.

Proof

We have verified the unsatisfiability of $\Phi$ via SAT solver. $\square$

The SAT solver runs into computational issues on $n \ge 20$ points, so we add more clauses to limit the search space. $\Phi _4$ and $\Phi _5$ assert that every light blue edge and every dark blue edge in Figure 3 has its needs satisfied, respectively, and $\Phi _6$ asserts that every edge that is not pre-colored has its needs satisfied: Define $BN = \{(0,0), (0,1), (0,2), (1,0), (2,0)\}$.

$$\begin{aligned} \Phi _4= & {} \bigwedge _{(i,j)\in Bl}\left[ \bigwedge _{(c_1,c_2)\in BN} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \\ \Phi _5= & {} \bigwedge _{(i,j)\in Bd}\left[ \bigwedge _{(c_1,c_2)\in BN} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \\ \Phi _6= & {} \bigwedge _{(i,j)\notin R\cup Bl \cup Bd } \phi _{i,j,0} \wedge \left[ \bigwedge _{(c_1,c_2)\in \{0,1,2\}} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \\&\vee \ \phi _{i,j,1} \wedge \left[ \bigwedge _{(c_1,c_2)\in BN} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \\&\vee \ \phi _{i,j,2} \wedge \left[ \bigwedge _{(c_1,c_2)\in BN} \left( \bigvee _{i\ne k \ne j} \phi _{i,k,c_1}\wedge \phi _{k,j,c_2}\right) \right] \end{aligned}$$

Lemma 4.4

Let $\Phi = \Phi _0 \wedge \Phi _1 \wedge \Phi _2\wedge \Phi _3\wedge \Phi _4 \wedge \Phi _5\wedge \Phi _6$. Then for $n\le 25$, $\Phi$ is unsatisfiable. Hence $20,21,22,23,24,25\not \in {{\,\mathrm{Spec}\,}}(32_{65})$.

Proof

We have verified the unsatisfiability of $\Phi$ via SAT solver. $\square$

Remark 4.5

We note that the Ramsey number $R(7, 3) = 23$. As $23 \not \in {{\,\mathrm{Spec}\,}}(32_{65})$, we obtain the following.

Corollary 4.6

In any representation of $32_{65}$, the clique number of the red subgraph of the underlying graph of the representation is at least seven.

5 Summary and open problems

We summarize our work as follows.

Theorem 5.1

We have $26 \le f(2) \le 1024$, and:

• (1) $f(n) \ge 2n^2+4n+1$ for all n.

• (2) $f(n) \le 2n^{6 + o(1)}$.

Problem 1

Is $f(2) < 1000$?

Problem 2

Is $32_{65}$ representable over $(\mathbb {Z}/2\mathbb {Z})^m$ for $m<10$? The natural thing to try—using the construction from the proof of Theorem 2.1, with $k=2$ (hence $m=7$)—doesn’t work; we checked all partitions. But there may be some other representation.

Problem 3

Can some modification of the technique used in [2] give a smaller representation of $32_{65}$? The most obvious thing to try—replacing ${[14] \atopwithdelims ()7}$ by ${[11] \atopwithdelims ()6}$—doesn’t work.

Problem 4

Which has the smaller minimal representation, $31_{37}$ or $32_{65}$? While $32_{65}$ has atoms $r, b_1, b_2$, all symmetric, with all-blue triangles forbidden, $31_{37}$ has atoms $r, b, b\,\breve{}$, with all-blue triangles forbidden. The atom r is flexible in both cases. The lower bound proven in Theorem 3.5 applies to representations of $31_{37}$ as well. The only (small) finite representation known to the authors is over $\mathbb {Z}/33791\mathbb {Z}$.